I have actually no idea how to proof it. I tried to use the rank-nullity theorem.F: field

V,W: F vector spaces

V*,W*: dual spaces (Hom(V,F);Hom(W,F))

φ : V -> W linear map

φ *:W*->V*

(λ: W -> F) |-> (λ∘ φ: V -> F)

Proof that dim(im(φ))=dim(im(φ*))

We know:

dim(V*)=dim(V)*dim(F)=dim(V)

dim(W*)=dim(W)*dim(F)=dim(W)

dim(V)=dim(ker(φ))+dim(im(φ))

dim(W)=dim(ker(φ*))+dim(im(φ*))

(r-n t)

Additionally:

dim(im(φ))<=dim(W)

dim(im(φ*))<=dim(V)

But how can I use these to show that the images have the same dimension? Maybe it is not possible how I tried it.

Is it somehow possible to show that there is an isomorphism between im(φ) and im(φ*)?

Has it something to do with the base of a dual space?